Project 2: Race Car Cornering

Part 1 - A Brief History of Auto-Racing

Car racing’s journey from rickety, steam-belching contraptions to today’s high-tech, world-famous supercars is packed with wild stories and weird traditions. It all kicked off in 1894 with the Paris-Rouen race, essentially a dare cooked up by a French newspaper to prove that these “horseless carriages” could actually go somewhere. Back then, “racing” was less about speed and more about keeping your vehicle in one piece — cars broke down, got lost, and generally limped to the finish. But people loved the spectacle, and car racing was born.

Soon, it wasn’t just local inventors; major companies joined in. In 1900, the Gordon Bennett Cup launched, inviting different nations to compete — the automotive Olympics of its time. National pride was on the line, and it was fierce. This competition laid the groundwork for some standard rules and introduced an international twist to the racing scene. Then, in 1906, we got the first true “Grand Prix” in France, where racers zoomed around closed circuits instead of weaving through the streets (and crowds), a move that saved some public panic and added to the event’s appeal.

With the 1910s came big names and epic distances, like the Indianapolis 500 in the U.S. This race gave drivers a chance to hammer down for 500 miles — at ridiculous speeds. In 1923, the 24 Hours of Le Mans took endurance to a new level, challenging racers to go non-stop for a full day and night. This legendary race tests not just speed but the mettle of both man and machine. By the time Formula One took off in 1950, car racing had become a global spectacle, with drivers achieving rockstar status and teams investing millions in shaving seconds off lap times.

Racetracks themselves became icons. The Monaco Grand Prix, weaving through city streets by the Mediterranean, is as glamorous as it is deadly, demanding precision and nerve. Silverstone, known as “the Home of British Motor Racing,” hosted the first Formula One race and has seen everything from spectacular wins to shocking crashes. The Nürburgring in Germany, nicknamed “The Green Hell” by legendary racer Jackie Stewart, is a track so twisty, challenging, and, frankly, terrifying that it’s in a league of its own.

In the 1960s and ’70s, racing entered a golden age of innovation, competition, and, yes, a little bit of craziness. The Ford-Ferrari rivalry became legendary as Ford took on Ferrari’s dominance at Le Mans. Formula One rivalries between teams like Ferrari, McLaren, and Lotus added drama on and off the track, fueling an era that left fans and media obsessed with the speed, danger, and glamour of racing.

Today, racing has grown beyond the screech of tires and roar of engines. Formula E is carving out a space for electric racing, while innovations in AI have introduced Roborace — racing, but with self-driving cars (think sci-fi meets Formula 1). Traditionalists may wince at the quiet hum of electric motors, but these changes are driving motorsport toward a greener, more futuristic path.

From rugged country roads to modern circuits packed with tech, car racing is a tale of competition, risk, and evolution — one that’s constantly fueled by innovation and a universal love for the thrill of speed.

My Sister at the Yas Marina Circuit

Part 2 - Navigating Flat Turns

The Walt Disney World Speedway in Orlando, Florida has three straightaways and three turns. The straightaways have lengths of $1487 \text{ft}$, $572 \text{ft}$, and $909 \text{ft}$. The turns have lengths of $990.35 \text{ft}$, $976.35 \text{ft}$, and $417.40 \text{ft}$.

I. Velocity and Acceleration Vectors

Let $\vec{r}_1(t)$ be the position vector of Babić’s race car at Turn 1:

$\qquad\vec{r}_1(t)=\begin{bmatrix} 446.85\cos(0.4759t) \\ 446.85\sin(0.4759t) \end{bmatrix}: t\in(0, \, 4.66)$

Differentiating with respect to $t$, we can get the velocity vector $\vec{v}_1(t)$:

$\qquad\vec{v}_1(t)=\vec{r}_{1}^{\prime}(t)=\begin{bmatrix} -212.6559\sin(0.4759t) \\ 212.6559\cos(0.4759t) \end{bmatrix}: t\in(0, \, 4.66)$$

Differentiating with respect to $t$ again, we can get the acceleration vector $\vec{a}_1(t)$:

$\qquad\vec{a}_1(t)=\vec{v}_{1}^{\prime}(t)=\vec{r}_{1}^{\prime\prime}(t)=\begin{bmatrix} -101.2029\cos(0.4759t) \\ -101.2029\sin(0.4759t) \end{bmatrix}: t\in(0, \, 4.66)$

To find speed $v_1$, we can simply take the magnitude of the velocity vector:

$\qquad\begin{aligned} v_1 &= \|\vec{v}_1(t)\| \\~\\ &= \sqrt{\vec{v}_1(t) \cdot \vec{v}_1(t)} \\~\\ &= \sqrt{212.6559^2[\sin^2(0.4759t)+\cos^2(0.4759t)]} \\~\\ &= \sqrt{212.6559^2} \\~\\ &= 212.6559 \; (\text{ft/s}) \end{aligned}$

Doing the same thing to find the magnitude of acceleration:

$\qquad\begin{aligned} a_1 &= \|\vec{a}_1(t)\| \\~\\ &= 101.2029 \; (\text{ft/s}) \end{aligned}$

We can also graph $\vec{r}_1(t)$, $\vec{v}_1(t)$, and $\vec{a}_1(t)$ on the same plane:

Now let $\vec{r}_2(t)$ be the position vector of Babić’s race car at Turn 2:

$\qquad\vec{r}_2(t)=\begin{bmatrix} 522.47\cos(0.4211t) \\ 522.47\sin(0.4211t) \end{bmatrix}: t\in(0, \, 4.44)$

We can repeat the same processes we used for Turn 1 to find $\vec{v}_2(t)$, $\vec{a}_2(t)$, and $v_2$:

$\qquad\begin{aligned} &\vec{v}_2(t)=\begin{bmatrix} -220.0121\cos(0.4211t) \\ 220.0121\sin(0.4211t) \end{bmatrix}: t\in(0, \, 4.44) \\~\\ &\vec{a}_2(t)=\begin{bmatrix} -92.6471\cos(0.4211t) \\ -92.6471\sin(0.4211t) \end{bmatrix}: t\in(0, \, 4.44) \\~\\ &v_2=220.0121 \; (\text{ft/s}) \\~\\ &a_2=92.6471 \; (\text{ft/s}) \end{aligned}$

Graphing the three vectors on the same plane:

Lastly, let’s set $\vec{r}_3(t)$ be the position vector of Babić’s race car at Turn 3:

$\qquad\vec{r}_3(t)=\begin{bmatrix} 733.89\cos(0.3797t) \\ 733.89\sin(0.3797t) \end{bmatrix}: t\in(0, \, 1.50)$

Again, using the same processes to find $\vec{v}_3(t)$, $\vec{a}_3(t)$, and $v_3$:

$\qquad\begin{aligned} &\vec{v}_3(t)=\begin{bmatrix} -278.6580\cos(0.4211t) \\ 278.6580\sin(0.4211t) \end{bmatrix}: t\in(0, \, 1.50) \\~\\ &\vec{a}_3(t)=\begin{bmatrix} -105.8065\cos(0.4211t) \\ -105.8065\sin(0.4211t) \end{bmatrix}: t\in(0, \, 1.50) \\~\\ &v_3=278.6580 \; (\text{ft/s}) \\~\\ &a_3=105.8065 \; (\text{ft/s}) \end{aligned}$

And lastly, graphing the three vectors together:

As we can see, the speed $v$ of the three turns are constant throughout each turn, equaling the coefficients of the trigonometric functions of their respective velocity vectors $\vec{v}(t)$. Their velocity vectors, however, are not constant; they are rotating. This is the same for acceleration; the acceleration $a$ of the three turns are constant throughout each turn, equaling the coefficients of the trigonometric functions of their respective acceleration vectors $\vec{a}(t)$. Their acceleration vectors, however, are not constant; they are also rotating. In fact, the acceleration vectors are always orthogonal to the velocity vectors.

II. Taking Turns

To find the curvature of each turn, we need to first find the radius. Since each turn is an arc of a circle, we can look at the coefficient of the trigonometric functions of each position function $\vec{r}(t)$ to determine the radius of the circle. Thus, for the three turns, we can get the following values:

Turn	Radius (ft)
Table of Turn Radii
1	446.85
2	522.47
3	733.89

Using the formula for curvature $\kappa=\frac{1}{r}$, we can plug in the three values of radii for the three different values for curvature:

Turn	Curvature (ft^-1)
Table of Turn Curvature
1	0.002237887
2	0.001913985
3	0.001362602

If we were to judge a turn’s difficulty based off of curvature alone, then it’s pretty simple. The greater the curvature, the more difficult it is. Thus, based off of curvature alone, Turn 1 is the hardest, and Turn 3 is the easiest.

$\qquad \text{Difficulty}: \; \text{Turn 1} > \text{Turn 2} > \text{Turn 3}$

However, this is only considering curvature alone. What if the race cars have a set speed they have to traverse these turns in? Let’s say that for each of these turns, they have to follow the velocity vector $\vec{v}(t)$ from before.

So how should we compare each turn? The answer: $a_N$.

$a_N$ is the normal component of acceleration, which means it is the component of the acceleration vector that is orthogonal to the velocity vector. Why use this metric? Because it is what makes the cars work. The greater this acceleration is, the greater force the car has to exert, since $F\propto a$.

To calculate $a_N$, we can simply take the magnitude of the cross product of $\vec{a}(t)$ and $\widehat{T}(t)$, where $\widehat{T}(t)$ is the normalized version of $\vec{v}(t)$, or $\frac{\vec{v}(t)}{v}$. But wait! We’re in 2D, so how do we take the cross product? To do so, we can use the dimensionless equivalent of the cross product: the wedge product (or exterior product).

Let’s walk through Turn 1 together. First we find $\widehat{T_1}(t)$:

$\qquad\begin{aligned} \widehat{T_1}(t) &= \frac{\vec{v}_1(t)}{v_1} \\~\\ &= \frac{−212.6559\sin(0.4759t)\widehat{\imath}+212.6559\cos(0.4759t)\widehat{\jmath}}{212.6559} \\~\\ &= -\sin(0.4759t)\widehat{\imath}+\cos(0.4759t)\widehat{\jmath} \end{aligned}$

Now to find the wedge product $a_N^B$ using the associative and distributive properties:

$\qquad\begin{aligned} a_N^B &= \vec{a}_1(t)\wedge\widehat{T_1}(t) \\~\\ &= (−101.2029\cos(0.4759t)\widehat{\imath}−101.2029\sin(0.4759t)\widehat{\jmath}) \wedge (-\sin(0.4759t)\widehat{\imath}+\cos(0.4759t)\widehat{\jmath}) \\~\\ &= −101.2029\cos(0.4759t)\widehat{\imath} \wedge \left[-\sin(0.4759t)\widehat{\imath}+\cos(0.4759t)\widehat{\jmath}\right] \\ & \quad\,−101.2029\sin(0.4759t)\widehat{\jmath} \wedge \left[-\sin(0.4759t)\widehat{\imath}+\cos(0.4759t)\widehat{\jmath}\right] \\~\\ &= 101.2029\cos(0.4759t)\sin(0.4759t)\widehat{\imath}\wedge\widehat{\imath}-101.2029\cos^2(0.4759t)\widehat{\imath}\wedge\widehat{\jmath} + \\ & \quad\, 101.2029\sin^2(0.4759t)\widehat{\jmath}\wedge\widehat{\imath}-101.2029\sin(0.4759t)\cos(0.4759t)\widehat{\jmath}\wedge\widehat{\imath} \end{aligned}$

Since $\widehat{\imath}\wedge\widehat{\imath}=0$, $\widehat{\jmath}\wedge\widehat{\jmath}=0$, and $\widehat{\imath}\wedge\widehat{\jmath}=-\widehat{\jmath}\wedge\widehat{\imath}$, we can simplify even further:

$\qquad\begin{aligned} a_{N1}^B &= -101.2029\cos^2(0.4759t)\widehat{\imath}\wedge\widehat{\jmath} - 101.2029\sin^2(0.4759t)\widehat{\imath}\wedge\widehat{\jmath} \\~\\ &= -101.2029\left[\cos^2(0.4759t)+\sin^2(0.4759t)\right]\widehat{\imath}\wedge\widehat{\jmath} \\~\\ &= -101.2029\widehat{\imath}\wedge\widehat{\jmath} \end{aligned}$

We can see that the wedge product of $\vec{a}_1(t)$ and $\widehat{T_1}(t)$ produces a bivector. To find the magnitude $a_{N1}$ of the bivector, we can use the formula $\sqrt{\langle a_{N1}^B(a_{N1}^B)^{\dagger} \rangle_0}$ (and the fact that $\widehat{\imath}^2=\widehat{\jmath}^2=1$):

$\qquad\begin{aligned} a_{N1} &= \sqrt{\langle a_{N1}^B(a_{N1}^B)^{\dagger} \rangle_0} \\~\\ &= \sqrt{\langle (-101.2029\widehat{\imath}\wedge\widehat{\jmath})(-101.2029\widehat{\imath}\wedge\widehat{\jmath})^{\dagger} \rangle_0} \\~\\ &= \sqrt{\langle 101.2029^2\left[(\widehat{\imath}\wedge\widehat{\jmath})(\widehat{\jmath}\wedge\widehat{\imath})\right] \rangle_0} \\~\\ &= \sqrt{\langle 101.2029^2(\widehat{\imath}\widehat{\jmath}\widehat{\jmath}\widehat{\imath}) \rangle_0} \\~\\ &= \sqrt{\langle 101.2029^2(\widehat{\imath}\widehat{\imath}) \rangle_0} \\~\\ &= \sqrt{\langle 101.2029^2\rangle_0} \\~\\ &= \sqrt{101.2029^2} \\~\\ &= 101.2029 \; (\text{ft/s}^2)\end{aligned}$

Since $a_{N1} = a_1$, we can conclude that for these circular turns, the magnitude of acceleration is the same as the normal component of acceleration (one can verify this by calculating the magnitude of the wedge product for the two other turns). Thus, to compare the normal component of acceleration of the turns, we can just compare their magnitude of acceleration:

Turn	Normal Acceleration a_N=a (ft/s²)
Table of Turn Normal Acceleration
1	101.2029
2	92.6471
3	105.8065

Looking at the table, it might come as a surprise that when taking speed into account, Turn 3 is actually the most difficult! Since it has the greatest normal component of acceleration, the car has to exert more force to follow the path with the given speed from before. By this logic, Turn 2 is still the easiest.

$\qquad \text{Difficulty}: \; \text{Turn 3} > \text{Turn 1} > \text{Turn 2}$

III. Graphing Velocity and Acceleration Vectors

Below is a table of values for $\vec{r}_1(t)$:

t	x(t)	y(t)
Table of r₁(t)
0.00	446.8500	0.0000
1.00	397.1964	204.7192
2.00	259.2704	363.9420
3.00	63.7246	442.2828
4.00	-145.9833	422.3314
4.66	-269.3227	356.5673

Here are the corresponding velocity vectors $\vec{v}_1(t)$ and acceleration vectors $\vec{a}_1(t)$ at each point of the table:

You can see that the velocity vectors $\vec{v}_1(t)$ are of equal length and tangential to the curve $\vec{r}_1(t)$, and the acceleration vectors $\vec{a}_1(t)$ are of equal length and orthogonal to the curve $\vec{r}_1(t)$, pointing towards the origin in the radial direction. This makes sense, since our conclusion from before was the the magnitude of acceleration is equal to its normal component, so the acceleration vector must be orthogonal to the velocity vector.

IV. Staying on Track

Usually to find the centripetal force required to stay on a circular path, we used the formula $F=m\frac{v^2}{r}$. However, since we already know the magnitude of acceleration for all three turns, we can just use $F=ma$. If the mass of Babić race car is $1800\text{lbs}$, we can calculate the force (in $\text{poundals}$, or $\text{ft/s}^2$) required for each turn:

Turn	Centripetal Force (pdl)
Table of Turn Force
1	182165.2
2	166764.8
3	190451.7

Part 3 - Driving Banked Turns

With the basics covered, we can now think more generally. Now we have to consider:

The weight of the car
The friction between the tires and the road
The radius of the osculating circle of the turn
The bank angle

The Bristol Motor Speedway in Tennessee has two straightaways and two turns. It is $0.533$ miles long, with bank angles that go all the way up to $30^{\circ}$. The track is mainly used for NASCAR race cars.

Let Babić’s race car of mass $m$ with a constant angular speed $\omega$ move around an arc of radius $R$ banked at an angle of $\theta$. Its center of mass off the ground by height $h$ and time $t$ is measured in seconds.

Let the car have a position function $\vec{r}(t)$:

$\qquad \vec{r}(t)=\begin{bmatrix} R\cos(\omega t) \\ R\sin(\omega t) \\ h \end{bmatrix}$

I. Velocity Function

To find the velocity function $\vec{v}(t)$ of the car, we simply take the derivative of $\vec{r}(t)$ with respect to $t$:

$\qquad \vec{v}(t)=\vec{r}^{\prime}(t) = \begin{bmatrix} -\omega R\sin(\omega t) \\ \omega R\cos(\omega t) \\ 0 \end{bmatrix}$

Since the tangent line of a circle is perpendicular to its radius, then if the dot product of $\vec{v}(t)$ and $\vec{r}(t)$ is $0$, then $\vec{v}(t)$ is tangent to the circular curve.

$\qquad\begin{aligned} \vec{v}(t)\cdot\vec{r}(t) &= \begin{bmatrix} -\omega R\sin(\omega t) \\ \omega R\cos(\omega t) \\ 0 \end{bmatrix} \cdot \begin{bmatrix} R\cos(\omega t) \\ R\sin(\omega t) \\ h \end{bmatrix} \\~\\ &= -\omega R^2 \sin(\omega t) \cos(\omega t) + \omega R^2 \cos(\omega t) \sin(\omega t) \\~\\ &=0 \end{aligned}$

Therefore the car’s velocity is tangent to the circular curve, and thus a centripetal force is needed to keep the car on the track.

II. Speed

To find speed $v$, we simply take the magnitude of the car’s velocity $\vec{v}(t)$:

$\qquad\begin{aligned} v &= \|\vec{v}(t)\| = \sqrt{\vec{v}(t)\cdot\vec{v}(t)} \\~\\ &= \sqrt{\omega ^2 R ^2 \left[\sin^2(\omega t) + \cos^2(\omega t)\right]} \\~\\ &= \sqrt{\omega ^2 R ^2} \\~\\ &= \omega R\end{aligned}$

Since $v=\omega R$, we can do some basic algebra do get the following equality:

$\qquad\begin{aligned} &v=\omega R \\~\\ &\therefore \frac{1}{v}=\frac{1}{\omega R} \\~\\ &\therefore \frac{R}{v}=\frac{1}{\omega} \\~\\ &\therefore \frac{2\pi R}{v}=\frac{2\pi}{\omega} \end{aligned}$

III. Acceleration

To find acceleration $\vec{a}(t)$, we simply take the derivative of the car’s velocity $\vec{v}(t)$:

$\qquad \vec{a}(t)=\vec{v}^{\prime}(t) = \begin{bmatrix} -\omega^2 R\cos(\omega t) \\ -\omega^2 R\sin(\omega t) \\ 0 \end{bmatrix}$

Now to find the magnitude of acceleration $a$:

$\qquad\begin{aligned} a &= \|\vec{a}(t)\| = \sqrt{\vec{a}(t)\cdot\vec{a}(t)} \\~\\ &= \sqrt{\omega ^4 R ^2 \left[\cos^2(\omega t) + \sin^2(\omega t)\right]} \\~\\ &= \sqrt{\omega ^4 R ^2} \\~\\ &= \omega^2 R\end{aligned}$

To see if $\vec{a}(t)$ points towards the center of the circle, we can take its cross product with $\vec{r}(t)$ with $h=0$. This way, both $\vec{a}(t)$ and $\vec{r}(t)$ are on the $xy-\text{plane}$, and the only way for their cross product to not be $\vec{0}$ is if they are not parallel. Thus, if the cross product equals to $\vec{0}$, then $\vec{a}(t)$ is indeed radial:

$\qquad\begin{aligned} \vec{a}(t)\times\vec{r}(t) &= \begin{bmatrix} -\omega^2 R\cos(\omega t) \\ -\omega^2 R\sin(\omega t) \\ 0 \end{bmatrix} \times \begin{bmatrix} R\cos(\omega t) \\ R\sin(\omega t) \\ h=0 \end{bmatrix} \\~\\ &= \begin{bmatrix} 0 \\ 0 \\ -\omega^2R^2\cos(\omega t)\sin(\omega)t+\omega^2R^2\cos(\omega t)\sin(\omega)t \end{bmatrix} \\~\\ &= \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} \\~\\ &=\vec{0} \end{aligned}$

Thus, the acceleration vector $\vec{a}(t)$ points towards the center of the circle.

IV. Centripetal Force

Let the centripetal force be $\vec{F}_c(t)$ and its magnitude be $F_c$. Since Newton's Second Law dictates that $\vec{F}=m\vec{a}$, we have the following three scalar equations:

$\qquad\begin{cases} F_c=ma \\ a=\omega^2R \\ \frac{2\pi R}{v}=\frac{2\pi}{\omega} \end{cases}$

To solve for $F_c$ in terms of $m$, $v$, and $R$, we can do some simple algebra:

$\qquad\begin{aligned} &\frac{2\pi R}{v}=\frac{2\pi}{\omega} \\~\\ &\therefore \frac{R}{v}=\frac{1}{\omega} \\~\\ &\therefore \frac{v}{R}=\omega \end{aligned}$

Now substitute $\omega$ in Equation 2:

$\qquad\begin{cases} F_c=ma \\ a=(\frac{v}{R})^2R=\frac{v^2}{R} \\ \frac{v}{R}=\omega \end{cases}$

Now we can substitute $a$ in Equation 1:

$\qquad F_c=ma \therefore F_c=m\frac{v^2}{R}$

V. Normal Force

The force that the ground exhibits on the car is perpendicular to the track, and is called the normal force, denoted by $\vec{N}$ with magnitude $N$. As for friction, it is denoted by $\vec{f}$ with magnitude $f$. The relationship between friction and the normal force is approximated by this equation:

$\qquad f=\mu N$

$\mu$ is a constant called the coefficient of friction, and is dependent on the condition of the track and the car’s tires.

The force of gravity can be described by $mg$, where $g$ is the gravitational acceleration at the surface of the Earth, which is around $32.1740 \text{ft/s}^2$.

These three forces can be drawn on Babić race car like so:

We can abstract these forces and take their vertical components:

To be precise, using basic trigonometry and vector projection (proof can be found using similar triangles), we can find that:

$\qquad \begin{cases} N_z=N\cos\theta \\ f_z=f\sin\theta \\ (mg)_z=mg \end{cases}$

Since the race car’s vertical acceleration (recall that the z-component of $\vec{a}(t)$ is always $0$), we know that all three of these vertical components of forces must be sum to $0$. Therefore, we can create an equation (noting that the direction of the component corresponds to their sign):

$\qquad N_z-f_z-(mg)_z=0$

With substitution, we can get:

$\qquad N\cos\theta-f\sin\theta-mg=0$

We can then rearrange to get:

$\qquad N\cos\theta=f\sin\theta+mg$

If we want to remove friction from the equation, then we can use the equation $f=\mu N$ to substitute:

$\qquad\begin{aligned} &N\cos\theta=f\sin\theta+mg \\~\\ &\therefore N\cos\theta=\mu N \sin \theta + mg \\~\\ &\therefore N(\cos\theta - \mu\sin\theta)=mg \\~\\ &\therefore N=\frac{mg}{\cos\theta - \mu\sin\theta} \end{aligned}$

VI. Centripetal Force Part 2

Similar to what we just did, instead of summing the vertical components of the three forces, we can sum the horizontal components.

Again, using trigonometry and vector projection, we can find that:

$\qquad \begin{cases} N_x=N\sin\theta \\ f_x=f\cos\theta \\ (mg)_x=0 \end{cases}$

Because the magnitude of centripetal force $F_c$ is just the sum of all the horizontal components, we can list the following equation and derive:

$\qquad\begin{aligned} &F_c=N_x+f_x+(mg)_x &\therefore F_c=N\sin\theta+f\cos\theta\end{aligned}$

But wait! We already have expressions for $N$ and $f$ in terms of $\mu$. Let’s substitute that in:

$\qquad\begin{aligned}&\begin{cases} F_c=N\sin\theta+f\cos\theta \\ f=\mu N \\ N=\frac{mg}{\cos\theta - \mu\sin\theta} \end{cases} \\~\\ &\therefore F_c=N\sin\theta+\mu N\cos\theta \\~\\ &\therefore F_c=N(\sin\theta+\mu\cos\theta) \\~\\ &\therefore F_c=\frac{mg}{\cos\theta-\mu\sin\theta}\cdot(\sin\theta+\mu\cos\theta) \\~\\ &\therefore F_c=\frac{\sin\theta+\mu\cos\theta}{\cos\theta-\mu\sin\theta}\cdot mg \end{aligned}$

VII. Maximum Speed

Now we have two equations for centripetal force:

$\qquad\begin{cases} F_c=m\frac{v^2}{R} \\ F_c=\frac{\sin\theta+\mu\cos\theta}{\cos\theta-\mu\sin\theta}\cdot mg \end{cases}$

Since second equation dictates the maximum centripetal force the car can have, the first equation limits the maximum speed the car can have before flying off the track. We can substitute out $F_c$ to find this maximum speed $v_{\text{max}}$:

$\qquad\begin{aligned} &m\frac{v^2}{R}=\frac{\sin\theta+\mu\cos\theta}{\cos\theta-\mu\sin\theta}\cdot mg \\~\\ &\therefore v^2= \frac{\sin\theta+\mu\cos\theta}{\cos\theta-\mu\sin\theta}\cdot gR \end{aligned}$

Because the mass of the car $m$ is not in the equation for $v_{\text{max}}$, we can conclude that the mass of the car actually does not affect the maximum speed it can go through a turn with.

VIII. Turning Sharp

Let’s take a look back at the Bristol Motor Speedway. The inside track of Turn 1 has a radius of $211\text{ft}$ and a bank angle of $24^{\circ}$. If we assume dry conditions so that the coefficient of friction $\mu=0.98$, we can calculate the maximum speed Babić’s race car can turn at:

$\qquad\begin{aligned} v_{\text{max}}^2 &= \frac{\sin\theta+\mu\cos\theta}{\cos\theta-\mu\sin\theta}\cdot gR \\~\\ &= \frac{\sin24^{\circ}+0.98\cos24^{\circ}}{\cos24^{\circ}-0.98\sin24^{\circ}}\cdot 32\cdot211 \\~\\ &= 17072.1230 \end{aligned} \\~\\ \qquad \therefore v_{\text{max}}=\sqrt{17072.1230}=130.6603 \; (\text{ft/s})$

But this is in feet per second. Let’s convert this to miles per hour. Doing so, we can conclude that in dry conditions, that maximum speed for a sharp turn is $89.0866\text{mph}$.

IV. Turning Wide

In contrast, the outside track of Turn 1 has a radius of $251\text{ft}$ and a bank angle of $28^{\circ}$. If we still assume dry conditions, we can calculate the maximum speed Babić’s race car can turn at:

$\qquad\begin{aligned} v_{\text{max}}^2 &= \frac{\sin\theta+\mu\cos\theta}{\cos\theta-\mu\sin\theta}\cdot gR \\~\\ &= \frac{\sin28^{\circ}+0.98\cos28^{\circ}}{\cos28^{\circ}-0.98\sin28^{\circ}}\cdot 32\cdot251 \\~\\ &= 25352.7302 \end{aligned} \\~\\ \qquad \therefore v_{\text{max}}=\sqrt{25352.7302}=159.2254 \; (\text{ft/s})$

Converting this to miles per hour, we get that in dry conditions, that maximum speed for a wide turn is $108.5628\text{mph}$.

X. Wet Turns

Back on the inside track of Turn 1, we’re now facing wet conditions so that the coefficient of friction is now as low as $\mu=0.1$. Now, we calculate that the maximum speed Babić’s race car can turn at:

$\qquad\begin{aligned} v_{\text{max}}^2 &= \frac{\sin\theta+\mu\cos\theta}{\cos\theta-\mu\sin\theta}\cdot gR \\~\\ &= \frac{\sin24^{\circ}+0.1\cos24^{\circ}}{\cos24^{\circ}-0.1\sin24^{\circ}}\cdot 32\cdot211 \\~\\ &= 3852.9275 \end{aligned} \\~\\ \qquad \therefore v_{\text{max}}=\sqrt{3852.9275}=62.0720 \; (\text{ft/s})$

Convert this to miles per hour, we can conclude that in wet conditions, that maximum speed for a sharp turn is $42.3218\text{mph}$.

XI. Coefficient of Friction

Now let’s suppose that Babić’s race car is going around Turn 1 at $105\text{mph}$ ($154\text{ft/s}$) on the outside track. We can then estimate her car’s tires’ coefficient of friction:

$\qquad\begin{aligned} &v_{\text{max}}^2 = \frac{\sin\theta+\mu\cos\theta}{\cos\theta-\mu\sin\theta}\cdot gR \\~\\ &\therefore 154^2=\frac{\sin28^{\circ}+\mu\cos28^{\circ}}{\cos28^{\circ}-\mu\sin28^{\circ}}\cdot 32\cdot251 \\~\\ &\therefore 23716=\frac{0.4695+0.8829\mu}{0.8829-0.4695\mu}\cdot 8032 \\~\\ &\therefore \mu=\frac{17167.8324}{18226.1148}=0.9419 \end{aligned}$

Thus, we can conclude that Babić’s car’s tires have a coefficient of friction of $\mu=0.9419$.

Bibliography

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